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Is it true or false that whenever $E \subseteq \mathbb{R}$ is such that $m^*(E) > 0$, where $$ m^*(E) = \inf\left\{\sum_{n=1}^\infty|b_n - a_n|\, :\mid\, E \subseteq \bigcup_{n = 1}^\infty(a_n,b_n)\right\} $$ there is some Borel set $B \subseteq E$ such that $m(B) > 0$?

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    $\begingroup$ Is $m_*$ supposed to be $m^*$? $\endgroup$ – user940 Oct 2 '14 at 15:15
  • $\begingroup$ @ByronSchmuland: Yes, yes it does. Corrected. Thanks. $\endgroup$ – Evan Aad Oct 2 '14 at 15:19
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Without assuming that $E$ is measurable, this is false.

Let $E$ be a Vitali set in $[0,1]$, so that if $\oplus$ denotes addition mod 1 and $\{q_i\}$ is an enumeration of the rationals in $[0,1)$, we have that $\{E \oplus q_i\}$ are pairwise disjoint and their union is $[0,1]$. I claim $m^*(E) > 0$. For by countable subadditivity of $m^*$ we have $$1 = m^*([0,1]) = m^*\left(\bigcup_{i=1}^\infty E \oplus q_i\right) \le \sum_{i=1}^\infty m^*(E \oplus q_i) = \sum_i m^*(E).$$ If $m^*(E) = 0$ then the right side would be 0.

Now suppose $B \subset E$ is Borel (or even just measurable). Then we have that $\{B \oplus q_i\}$ are pairwise disjoint. By monotonicity and countable additivity of $m$ we have $$1 = m([0,1]) \ge m\left(\bigcup_{i=1}^\infty B \oplus q_i\right) = \sum_{i=1}^\infty m(B \oplus q_i) = \sum_{i=1}^\infty m(B).$$ If $m(B) > 0$ the right side would be $\infty$. So we must have $m(B)=0$ and hence $m^*(B)=0$.

If $E$ is assumed to be measurable, this is true. Indeed, you can find a Borel set $B \subset E$ with $m(B) = m(E)$. There should be a proof elsewhere on this site, but I cannot find one right now.

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It is true. Given any measurable set $E$ in $\Bbb R$, given $\epsilon > 0$, exists $F \subset E$ closed such that ${\frak m}^*E < {\frak m}^*F + \epsilon$. Given any measurable set $E$ in $\Bbb R$, exists $F \in F_\sigma$ contained in $E$, such that ${\frak m}^*(E \setminus F) = 0$.

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  • $\begingroup$ Thanks. Where can I find a proof of this result, please? $\endgroup$ – Evan Aad Oct 2 '14 at 12:19
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    $\begingroup$ It is a proposition in Royden's Real Analysis book, in Lebesgue Measure chapter, but I don't remember right now if he proves it or if he gives hints and leave it as an exercise. But it is not hard. You first prove an analogous result with open sets and $G_\delta$s, and this one goes as a corollary by taking complements. $\endgroup$ – Ivo Terek Oct 2 '14 at 12:26
  • $\begingroup$ @EvanAad take a look at this. $\endgroup$ – Ivo Terek Oct 2 '14 at 14:16
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    $\begingroup$ The question doesn't say anything about $E$ being measurable. $\endgroup$ – user940 Oct 2 '14 at 15:16
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The question of Evan Aad is not solvable within the theory $ZF+DC$. Indeed, in the theory $ZFC$ which is a consistent extension of $ZF+DC$(of course, if $ZF+DC$ is consistent) the answer is no. See, for example the answer of Nate Eldredge. On the other sense, in Solovay Model (ZF+DC+"every subset of real axis is Lebesgue measurable") which also is a consistent extension of $ZF+DC$(of course, if $ZF+DC$ is consistent) the answer is yes. Indeed, in Solovay Model each subset with positive Lebesgue outer measure same times is Lebesgue measurable with positive Lebesgue measure. Now by virtue an inner regularity of the Lebesgue measure one can find a required Borel subset.

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